Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9216 Accepted Submission(s): 4220
Problem Description
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M].
If there are more than one K exist, output the smallest one.
Input
The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate
a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
Output
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input
2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1
Sample Output
6 -1
模板题，代码：
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<time.h>
#include<vector>
using namespace std;
typedef long long ll;
ll a[1000010],b[1000010];
int next[1000010],m,n;
void getnext()
{
int i,j,k;
next[0]=-1;
j=0,k=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
{
j++;
k++;
next[j]=k;
}
else k=next[k];
}
}
int kmp()
{
int i=0,j=0;
getnext();
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else j=next[j];
}
if(j==m)return i-m+1;
else return -1;
}
int main()
{
int T,i,j,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)scanf("%d",&a[i]);
for(i=0;i<m;i++)scanf("%d",&b[i]);
printf("%d\n",kmp());
}
}